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6. (a) A particle of mass m kg is suspended from a fixed point p by a light elastic string - Leaving Cert Applied Maths - Question 6 - 2007
Question 6
6. (a) A particle of mass m kg is suspended from a fixed point p by a light elastic string. The extension of the string is d when the particle is in equilibrium. T... show full transcript
Worked Solution & Example Answer:6. (a) A particle of mass m kg is suspended from a fixed point p by a light elastic string - Leaving Cert Applied Maths - Question 6 - 2007
Step 1
Show that the motion of the particle is simple harmonic.
Answer
At equilibrium position, the extension x is 0; thus, the tension in the string equals the weight of the particle:
[ T = kd = mg ]
As the particle is displaced a distance x (where 0 < x < d), the new position's forces can be determined:
[ T = mg - k(d + x) ]
This simplifies to:
[ mg - kd - kx = 0 ]
Substituting T with negative x:
[ -kx = -mg + kd ]
From the above, we can deduce the net force acting on the particle:
[ F = -kx ]
Since this is the classical form of Hooke's Law, the particle exhibits simple harmonic motion (S.H.M) with an angular frequency:
[ \omega = \sqrt{\frac{k}{m}} \]
Step 2
Find, in terms of d, the period of the motion.
Answer
The period T of simple harmonic motion can be derived from the angular frequency:
[ T = \frac{2\pi}{\omega} ]
Substituting the expression for ( \omega ):
[ T = \frac{2\pi}{\sqrt{\frac{k}{m}}} ]
To express this in terms of d, we recognize that:
[ k = \frac{mg}{d} ]
Thus:
[ \sqrt{\frac{k}{m}} = \sqrt{\frac{g}{d}} \]
By substituting back to find T:
[ T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{m\cdot d}{mg}} = 2\pi \sqrt{\frac{d}{g}} ]
Step 3
Show that the combined mass will not reach the point c.
Answer
Let ( v ) be the velocity of c when it reaches d. At point d, considering total energy: [ \text{Total energy at c} = \text{Total energy at d} ] Using conservation of energy, we express: [ \frac{1}{2} m (\sqrt{10gr})^2 + mg(2r) = \frac{1}{2} m v^2 + mg(0) ] This simplifies to: [ 14gr = v^2 ] Denote the combined mass as ( 2m ) when reaching point d. Using ( \frac{1}{2} (2m) v^2 + (2m)g(2r) = \frac{1}{2} (2m) v_2^2 + (2m)g(0) ] We conclude: [ 14gr = v_2^2 + 4gr ] Resulting in: [ v_2^2 = 14gr - 4gr = 10gr ] This leads to: [ v_2 = \sqrt{10gr} ] Since we need to equalize time and energy at point c, observe that: [ v_2 < \sqrt{14gr} ] Thus, the combined mass will not possess sufficient energy to ascend back to c after coalescence.